I just started programming so pointers and arrays confuse me.
This program just assigns random numbers from 0 - 9 into array and prints them out
(#include <stdio.h> #include <stdlib.h> #include <time.h>)
int function(int *num[]){
int i;
for(i=0; i<10; i++){
srand((unsigned)time(NULL));
*num[i] = rand()%10;
printf("%d", *num[i]);
}
return 0;
}
int main(){
int num[10];
function(&num); // incompatable pointer type (how do i fix this?)
return 0;
}
Thankyou
Change your code to read something like this:
int function(int num[]){
int i;
for(i=0; i<10; i++){
srand((unsigned)time(NULL));
num[i] = rand()%10;
printf("%d", num[i]);
}
return 0;
}
int main(){
int num[10];
function(num);
return 0;
}
In main(), you are allocating an array of 10 integers. The call to function(num) passes the address of this array (technically, the address of the first element of the array) to the function() function. In a parameter declaration, int num[] is almost exactly equivalent to int *num (I'll let you ponder on that one). Finally, when you access the elements of num[] from within the function, you don't need any extra *. By using num[i], you are accessing the i'th element of the array pointed to by num.
It may help to remember that in C, the following are exactly the same:
num[i]
*(num+i)
You don't need to pass in a pointer to your array. Just pass in the array. I have fixed your pointer code below.
Also, you should not reset the seed (srand) through every iteration through the for loop.
int function(int num[]){
int i;
srand((unsigned(time(NULL));
for(i=0; i<10; i++){
num[i] = rand()%10;
printf("%d", num[i]);
}
return 0;
}
int main(){
int num[10];
function(num);
return 0;
}
It all depends on what you want to achieve in the end.
(1) If your function function is intended to be used with arrays of run-time size, then you have to pass a pointer to the first element of the array. That is achieved by the following equivalent declarations
int function(int *num)
or
int function(int num[])
Inside the function you have to access the passed array as
num[i] = rand() % 10;
And you call your function as
function(num);
Of course, when the array size is a run-time value, it makes sense to pass that size to the function as well, meaning that your function should have the following interface
int function(int num[], size_t n)
And implement it for an array of size n, instead of hardcoding 10 directly into your implementation.
(2) If your function function is intended to be used with arrays of fixed compile-time size (10 in this case), then the better approach would be to pass a pointer to the entire array. It is achieved by the following declaration
int function(int (*num)[10])
Inside the function you have to access the passed array as
(*num)[i] = rand() % 10;
And you call your function as
function(&num);
So it is either (1) or (2).
What you currently have in your code looks like a mix of these two approaches. More precisely, you code looks like an attempt to implement the second approach, but it is missing some important parentheses :)
Most answers given to you so far suggest using the first approach. However, seeing that the array size is actually a compile-time constant in your case, I would suggest sticking with the second approach.
Use either int *num or int num[], don't use both together as you have right now (int *num[]). An array is passed to a function as a pointer to the first element, i.e. int *num, which is equivalent to int num[].
Related
I will go straight to what I'm asking for, I also see some similar question but is not what I'm looking for...so it seems I have to ask with a new forum.
I'm preparing myself for a future examination, where is not required the pointer, but I would like to get some extra information and abilities.
Here's the code followed by the question...
I'm using Fedora 33, I know is different from some IDE on Windows (ex: Visual Studio or Dev C++)
/* It's just a simple test, if this work I will get myself into a more complicated one, as you could read in the
* forum, I'm getting ready ( just a recheck of my abilities ) for an universitary examinaton. */
#include <stdio.h>
#include <stdlib.h>
#define N 5
void casual_generation(int** mat);
void prompt_print(int** mat);
int main()
{
int **mat[N][N];
casual_generation(**mat);
prompt_print(**mat);
}
void casual_generation(int** mat)
{
int i=0,j=0;
for(i=0;i<N;i++)
for(j=0;j<N;j++)
mat[i][j] = rand() % 50;
}
void prompt_print(int** mat)
{
int i=0,j=0;
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
printf("%d ", mat[i][j]);
printf("\n");
}
}
Somebody else on the forum used malloc, struct or other stuff, as you can see in this picture, when I try to execute him it says "Segmentation fault (core dumped)"
screen error
Where is my error?
And if you want, can you also send me the version with the passed value pointer?
Thanks for whoever will give me an answer, and time dedicated.
This declaration
int **mat[N][N];
does not make a sense. It means that you have a matrix elements of which are pointers of the type int **. But you need a matrix elements of which are integer numbers of the type int. That is you need a declaration like this
int mat[N][N];
So now you have a two-dimensional array (or matrix) of integers.
As you are going to pass this two-dimensional array to functions then used as an argument expression it is converted to pointer to its first element of the type int ( * )[N].
Correspondingly the functions that accepts such an array should be declared like
void casual_generation( int mat[][N], size_t n );
void prompt_print( int mat[][N], size_t n );
or (that is fully equivalent) like
void casual_generation( int ( * mat )[N], size_t n );
void prompt_print( int ( *mat )[N], size_t n );
because the compiler adjusts function parameters having array types to pointers to array element types.
Now for example the first function can be defined the following way
void casual_generation( int ( * mat )[N], size_t n )
{
for ( size_t i = 0; i < n; i++ )
{
for ( size_t j = 0; j < N; j++ )
{
mat[i][j] = rand() % 50;
}
}
}
And the function can be called like
casual_generation( mat, N );
A similar way can be defined the function prompt_print.
Using the second parameter makes the function more general. For example it can be called for two-dimensional arrays with different numbers of rows.
Now I will explain why you are getting a segmentation fault in your original code.
You have this declaration
int **mat[N][N];
a two dimensional array of pointers of the type int **.
Then you are using the expression **mat as an argument of function calls like this
casual_generation(**mat);
Then you are applying the dereference operator like *mat the array designator is converted to pointer to its first element (row) having the type int ** ( * )[N]. So dereferencing this pointer you get the first row of your array int **[N]. Applying the second time the dereferenced operator to this expression that has an array type the used expression is again is converted to pointer to its first element of the type int **( * ). That is it points to the first element of the first row of the original two-dimensional array. Dereferencing this pointer you get the first element of the type int **. This uninitialized pointer with indeterminate value the function accepts as its argument.
Thus dereferencing this first uninitialized element of the original matrix within the function
mat[i][j] = rand() % 50;
^^^
you get a segmentation fault. The reason of the fault is the incorrect matrix and the corresponding function parameter as it was shown above in tbe beginning of the answer.
Where is my error?
The "Segmentation fault" error happens because you define the variable mat as a pointer, but don't allocate any memory for it to point to.
int **mat[N][N];
You meant to do
Int mat[N][N];
and
casual_generation(mat);
prompt_print(mat);
By passing **mat you are passing mat[0][0] that is an int, but you want to pass the whole matrix which is a pointer to pointers to int (i.e. int **)
And you may want to introduce srand() in your code.
Just to make things clear:
mat is of type int ** and it's the whole matrix (or if you want it's a pointer to the first row)
*mat is of type int * and it's the first row of the matrix (or if you want it's a pointer to the first element of the first row)
**mat is of type int and it's the first element of the first row of the matrix
int **mat[N][N];
Here, you defined a double pointer to a 2D array. You only need to use one of those - a double pointer or a 2D array, like so:
int mat[N][N];
However, the bigger problem comes from trying to interchange 2D arrays and double pointers. This isn't possible in C since the 2D array is laid out flat in memory.
You need to create an array mat_ptr of pointers yourself and then pass that to casual_generation and prompt_print.
Finally, these casual_generation and prompt_print functions expect to be given a pointer, so you shouldn't dereference the pointer with ** before calling the function.
The final working code is:
int main()
{
int mat[N][N];
int *mat_ptr[N];
for (int i = 0; i < N; i++)
mat_ptr[i] = mat[i];
casual_generation(mat_ptr);
prompt_print(mat_ptr);
}
As you can find a detailed explanation why the code crashed in the other answer I will only propose a quite elegant solution that uses a neat though little known feature from C99 called Variable Length Arrays (aka VLA).
#include <stdio.h>
#include <stdlib.h>
void casual_generation(int n, int mat[n][n]);
void prompt_print(int n, int mat[n][n]);
int main()
{
const int N = 5;
int mat[N][N];
casual_generation(N, mat);
prompt_print(N, mat);
}
void casual_generation(int n, int mat[n][n])
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
mat[i][j] = rand() % 50;
}
void prompt_print(int n, int mat[n][n])
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
printf("%d ", mat[i][j]);
printf("\n");
}
}
It compiles in pedantic mode with no warnings and it works like a charm.
VLA were introduced to C to simplify numerical computation over multidimensional arrays.
#include <stdio.h>
void print(int (*p)[3]);
int main(void)
{
int a[3] = {1, 2, 3};
print(&a);
return 0;
}
void print(int (*p)[3])
{
for (int i = 0; i < sizeof(*p) / sizeof(**p); i++)
printf("%d\n", (*p)[i]);
}
I have written a C function. See above.
It can print all the elements in an array.
There is one thing which is not so perfect : The number of array elements seems to be known in advance.
So I made some modification in hopes of making the function universal :
#include <stdio.h>
void print(int (*p)[]);
int main(void)
{
int a[3] = {1, 2, 3};
print(&a);
return 0;
}
void print(int (*p)[])
{
for (int i = 0; i < sizeof(*p) / sizeof(**p); i++)
printf("%d\n", (*p)[i]);
}
In the function, p is pointer pointing to the entire array.
However, it fails to be compiled.
Why can't int (*p)[] be used as an argument for C function?
int (*p)[] can be used as an argument for the function. The part of your code that gives the error is sizeof *p which is obviously not possible, because the type of *p is int[] which is an incomplete type and hence has no known size.
For the function to know the length of the array, you must design a way for the function to receive this information. Options include:
what you did in the original code.
passing the length as another argument.
including the length as an array element.
having a sentinel value on the end of the array.
The most common idiom would be to pass int *p, size_t n, you do not really gain anything by using pointer-to-array without the dimension being given.
The problem is that int [] is an incomplete type as the array has no defined size and therefore its sizeof cannot be taken.
In "modern C" (i.e. for almost 2 decades) you could have used variable-length arrays for this - you can pass the size as an argument and then the array:
#include <stdio.h>
#include <stdlib.h>
void print(size_t n, int (*p)[*]);
int main(void) {
int a[3] = {1, 2, 3};
print(3, &a);
}
void print(size_t n, int (*p)[n]) {
for (size_t i = 0; i < sizeof(*p) / sizeof(**p); i++)
printf("%d\n", (*p)[i]);
}
Of course this gains you nothing, since sizeof *p / sizeof **pp will be ... n. Therefore we might as well have used
void print(size_t n, int p[n]) {
for (size_t i = 0; i < p; i++)
printf("%d\n", p[i]);
}
which is less typing.
Short answer: int (*p)[] can't be used as an argument and have the function magically know the array size, because the standard says so.
Longer answer:
int (*p)[] is a pointer to an array, but the array has no defined size. So by looking at the array, it is impossible to do pointer arithmetic, calculate size of the thing p is pointing at, etc..
You don't have array of arrays so you don't need int (*p)[]. You have an array of int, so int *p or int p[] should be enough. This does not solve the problem of knowing the size of your array in print. To do this you basically have 3 options
Hardcode the value in the function
Put a sentinel value in your array to mark the end
Pass the size as a separate parameter like this:
void print(int n, int p[n])
Just remember that whatever method you use, parameter passing of arrays will always use pointers behind the scenes, so you CAN NOT use sizeof(p) to calculate the size of the array. sizeof will always return the size of a pointer in those situations
I am recently studying pointers and arrays.
Suppose I initialize an array like
int a[4]={6,2,3,4};
Now after reading a lot ,I understand that
1) a and &a will point to same location.But they aren't pointing to the same type.
2) a points to the first element of the array which is an integer value so the type of a is int*.
3) &a points to an entire array(i.e an integer array of size 4) therefore the type is int (*)[].
Now what I actually don't get is how to use these type??
For Example:
CODE 1:
#include<stdio.h>
void foo(int (*arr)[4],int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",*(*arr+i));
printf("\n");
}
int main()
{
int arr[4]={6,2,3,4};
foo(&arr,4);
return 0;
}
CODE 2:
#include<stdio.h>
void foo(int *arr,int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",*(arr+i));
printf("\n");
}
int main()
{
int arr[4]={6,2,3,4};
foo(&arr,4);
return 0;
}
In code 2 we are passing &arr so its type should be of int (*)[],then how come we are getting the correct output even though we are having a type int *.
I really don't understand what is the meaning of type and how to use it?
Kindly explain with some examples.
Code 2 is not valid C, because a conversion between an array pointer and a pointer to int is not a valid pointer conversion - they are not compatible types. If it works, it is because of some non-standard extension of your compiler, or possibly you just "got lucky".
Please note that the best way to pass an array to a function in modern C is this:
void foo(int n, int arr[n])
{
for(int i=0;i<n;i++)
printf("%d ",arr[i]);
printf("\n");
}
As part of a function parameter list, arr will get adjusted to a pointer to the first element of the array, type int*.
The pedantically correct version would also replace int n and int i with size_t, which is the most proper type to use when describing the size of an object.
The best way to pass arrays in to functions is a pointer and the number of elements.
Note you can pass arr in to the function as arr or &arr[0]; both resolve to the same address.
#include<stdio.h>
void foo(int *arr,int n)
{
int i;
for(i=0;i<n;i++)
printf("%d ",arr[i]);
printf("\n");
}
int main()
{
int arr[4]={6,2,3,4};
foo(arr,4);
return 0;
}
Also, it's better to index the array, rather than addition and dereference. e.g. *(arr+i) becomes arr[i].
I have been learning arrays, but theres one thing that I cant figure out.
I borrowed two books for C and looked online, but found no solution.
My function timesTen multiplies every array elemenet that I have by 10,
then returns pointer of that array back function main()
How can I copy array a[2] directly in array x[2]?
I would usually use for loop, but I cant, because arguments are in two different functions.
Solution has probably got something to do with pointers, so feel free to post sollution here, but is there any way around them aswell?
Heres the source code:
#include <stdio.h>
int timesTen(int a[])
{
int i;
for (i=0;i<2;i++)
{
printf("%d\t", a[i]);
a[i]*=10;
printf("%d\n", a[i]);
}
return a;
}
int main()
{
int i;
int x[2];
int a[2]={10,50};
// i know here's an error, but how do I fix it? I cant put x[0]=timesTen(a[0])
x[2] = timesTen(a);
//also what if there is array a[10], and I want to copy it in x[5]
for (i=0;i<2;i++)
printf("%d\n", x[i]);
return 0;
}
Thanks!
What you need to understand is the distinction between arrays and pointers. When you declare your two arrays in main(), you allocate two times memory for two integers. That's fine. But in C, you simply cannot pass arrays around (as in: implicitly allocate a new slap of memory and copy the data of the source array into this memory region). Instead, any array identifier will decay to a pointer to the first element of the array in almost all situation. So when you write
int x[2];
int a[2]={10,50};
timesTen(a);
this code is precisely equivalent to
int x[2];
int a[2]={10,50};
timesTen(&a[0]);
So, why does that not clash with your declaration of timesTen()? Because array parameters in function declarations decay right there, on the spot, into a pointer! So, your function declaration is precisely equivalent to this one:
int timesTen(int* a) {
This is one of the least understood features of the C language, and admittedly, it is hard to wrap your brain around this, but once you understand what pointer decay means, you will be much more at ease using pointers and arrays.
So, back to your question. Since you passed only a pointer to your array to timesTen(), and since you modify this array, the changes are directly visible in main(). There are two ways to achieve the behavior you want:
You can change the definition of timesTen() to copy the data into a destination array:
void timesTen(int size, int* source, int* dest) {
for(int i = 0; i < size; i++) dest[i] = 10*source[i];
}
int main() {
int x[2];
int a[2]={10,50};
timesTen(2, a, x); //pointer decay!
//x now contains {100, 500}
}
You can copy the data into the destination array before calling your function to modify the destination array:
void timesTen(int size, int* data) {
for(int i = 0; i < size; i++) data[i] = 10*data[i];
}
int main() {
int x[2];
int a[2]={10,50};
memcpy(x, a, sizeof(a)); //the sizeof operator is one of only two places where no pointer decay happens!
timesTen(2, x); //pointer decay!
//x now contains {100, 500}
}
In the function timesTen, since a is an array, each modification made to it in the function is also done to the parameter you passed (call by address not by value). Therefore you don't need to returns anything.
void timesTen(int a[])
{
int i;
for (i=0;i<2;i++) a[i]*=10;
}
And you just call it by:
timesTen(a);
You probably want something like this:
timesTen(a);
memmove(x, a, 2 * sizeof(x[0]));
instead of
x[2] = timesTen(a);
Note that your function does not need to return anything, because it is modifying the array on its original place. In C if you have an array parameter, it means that only a pointer is passed, not the whole array.
in main function:
int *p;
int i;
p = timesTen(a);
for ( i = 0; i < 2; i++ )
{
printf( "%d\n",*(p + i)); // here you can print the values returned from your function
}
Through pointers you could have eaisly managed it
main ()
{
int a[ 2 ];
int *ptr = timesTen(a);
for ( int i=0; i<2 ; i++)
{
printf("%d",ptr[i]);
}
And as far as
x[2] = timesTen(a);
Is concerned note that x[2] will give "value at 2nd adress from adrees of base that is x"
And it is not a variable but it is a value and you cant assign to a value.
Technically x[2] is not a lvalue.
I had to write a program which has a function that sums all positive number in an array, but using function parameter array as an pointer. There in main function a problem occurs, when i try to call sum function. here is my source code:
#include <stdio.h>
#include <conio.h>
int posit(int *x[], int n){
int s=0;
for(int i=0; i<n; i++){
if(*x[i]>0){
s=s+*x[i];
}
}
return s;
}
int main(){
int k;
scanf("%d",&k);
int a[k];
for(int i=0; i<k; i++){
scanf("%d",&a[i]);
}
int b=posit(&a,k);
printf("%d\n", b);
getch();
return 0;
}
When an array is passed to a function it decays to a pointer to its first element. Change the signature of the function to:
int posit(int* x, int n){ /* Equivalent to 'int posit(int x[], int n)' */
and change the element access syntax of the array in the function from *x[i] to just x[i]. The invoke the function:
int b=posit(a,k);
Another point worth mentioning is to check the return value of scanf(), which returns the number of assignments made, to ensure &a[i] is actually being assigned and the program is subsequently not using uninitialized variables.
You may find the Arrays and Pointers section of the C FAQ useful.
Actually in your example you are passing to function pointer to pointer (please note that similar notation is very often used with main argument char *argv[]).
As mentioned in hmjd post while passing array to function this is converted to address of its first element. Instead of explicitly using pointer (int *x) you can also use array notation (int[] or even int[k]) but those are the same.
Please also note that in second method (int[k]) size provided doesn't matter and inside function it will be not known. This is only important to include size when you are dealing with multiple dimension arrays, but this is other story...
From K&R2
When an array name is passed to a function, what is passed is the
location of the initial element. Within the called function, this
argument is a local variable, and so an array name parameter is a
pointer, that is, a variable containing an address.
So you will be passing only the first element to the function.
Once this is understood your source can be modified as follows
int posit(int *x, int n)
and to call this function from main you can use
int b=posit(a,k); //Note only the location of the first element is being passed